Question

Draw the orientations of electric dipole moment in a uniform electric field such that the dipole experiences

A) minimum torque

B) half of the maximum torque

C) 1/2^1÷2 times the maximum torque

D) 76.6% of the maximum torque

Show the two possible ways of orientation since there are two possible angles in each case

Answer 1

To Draw:

Orientation of electric dipole in a uniform electric field such that the torque varies as follows :

• Minimum torque

• Half of max torque

• 1/√2 times of max torque

• 76.6% of max torque

Calculation:

Maximum torque is given as :

$\boxed{ \sf{ \tau = P \times E \times \sin(90 \degree) = PE}}$

1st condition:

Dipole experiences minimum torque ;

Minimum value of torque is zero , so we can say:

$\therefore \: \tau = 0 = PE \sin( \theta)$

$= > \sin( \theta) = 0$

$= > \theta = 0 \degree \: or \: 180 \degree$

2nd condition

Dipole experiences half of max torque ;

$\therefore \: \tau = PE \sin( \theta) = \dfrac{1}{2} PE$

$= > \sin( \theta) = \dfrac{1}{2}$

$= > \theta = 30 \degree \: or \: 150 \degree$

3rd condition:

Dipole experiences 1/√2 times of max torque :

$\therefore \: \tau = PE \sin( \theta) = \dfrac{1}{ \sqrt{2} } PE$

$= > \sin( \theta) = \dfrac{1}{ \sqrt{2} }$

$= > \theta = 45 \degree \: or \: 135\degree$

4th condition:

Dipole experiences 76.6% of max torque :

$\therefore \: \tau = PE \sin( \theta) = 76.6\% \times PE$

$= > \sin( \theta) = \dfrac{76.6}{ 100 }$

$= > \sin( \theta) = 0.766$

$= > \sin( \theta) = \dfrac{1.414}{2}$

$= > \sin( \theta) = \dfrac{ \sqrt{2} }{2}$

$= > \sin( \theta) = \dfrac{ 1}{ \sqrt{2} }$

$= > \theta = 45 \degree \: or \: 135\degree$

Now , refer to the attached diagram to understand the orientation of the dipoles.